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Part C find the particular solution W as a function of T to the differential equations derivative of W with respect to t is equal to 1 over 25 times W minus 300 with initial condition at time 0 W is equal to 1,400 and the units in this problem it was 1,400 tons they told us 1400 tons so it might seem strange to see a differential equations problem on an AP exam you say well you know I said this isn't fair this isn't you know this isn't a differential equations class and the thing to keep in mind is if on an AP exam you see differential equations they're really asking you to solve differential equations that can be solved using the skills that you would learn in an AP class and the the main differential or class of differential equations that you could solve are the ones are separable differential equations the ones where you can separate the independent and the dependent variables and if you don't have any idea what I'm talking about I'm about to show you so let me rewrite this differential equation we have DW DT is equal to 1 over 25 times W minus 300 now this differential equation we have W here and we have this DT but we don't have T anywhere else so what I'm what I want to do is I want to get the parts that involve W on to the left-hand side and I want to get anything that deals with DT on the right-hand side so let's let's divide both sides of this equation by W minus 300 so we would have 1 over W minus 300 DW DT the derivative of W with respect to T is equal to well I divided this side by W minus 300 so it's is equal to 1 over 25 now I want to get the DTS on this side so let's multiply both sides by DT viewed the differential is just a very very small change in T and so we get we get and I'm a little bit imprecise with my notation here but this is one way that you can think about it when you are dealing with the differential equations like this separable differential equations you have 1 over W minus 300 DW and make sure that I'm that the DW doesn't look like it's in the denominator DW is equal to is equal to 1 over 25 d and now we can integrate both sides which is essentially just saying that we're taking since this is equal to this that each little increment of W times this is equal to each little increment of DT times 1 over 25 we can then say well if we sum up all of the increments of W times this quantity if we sum up that the the infinite number of infinitesimally small increments that would be the same thing as summing up all of these characters because each of these characters is really the same as each of these characters so let's do that some we're essentially just taking we're just integrating both sides of this and here you might be able to do this in your head or we could do the formal use substitution just to make it clear what we're doing if you said u is equal to W minus 300 then what is D you if you wanted a differential so if you if you want to take the derivative of U with respect to W you would say D U with respect to W is equal to what it is equal to 1 the derivative of this with respect to W is 1 or if you multiply both sides times DW you have d U is equal to DW and so this left-hand side right over here this integral could be re-written as the integral of 1 over u D U we define this as U and we just saw that D U is the same thing as DW so this could be rewritten as d U and this is one of the basic integrals so hopefully we've learned and this is equal to the natural log of the absolute value of U or if you were if you unwind the substitution this is equal to the natural log of the absolute value of W minus 300 so let me write this down so the left-hand side right over here and you might be able do that in your head you say hey look you have W minus 300 its derivative is 1 so it's essentially there so I can just take this as if this was 1 over I can just anti take the antiderivative here and treat this whole thing is kind of a variable so it would just be the natural log of this entire thing or the absolute value of that entire thing so the integral the integral of the left hand side is the natural log of W minus 300 of the absolute value of W minus 300 and the the right-hand side is a little bit more straightforward with respect to T or sorry the inter the antiderivative with respect to T on the right hand side is 1 over 25 1 over 25 T or we could say T over 25 whereas T over 25 and then these are both indefinite integrals you could have a constant on both sides or one side but we could just put it on one side right over here you could have put plus some constant you could have added some you know c1 over here and then you could have had c2 over here but then you could have subtracted c1 from both sides and then just have a seat three constant over here but to simplify it you can just say look we just we this is this is going to be equal to this plus some constant you could put the constant on that side as well it doesn't matter all the constants would kind of merge into one now this is we've gotten pretty far we want to solve we want to solve the particular solution right now we only have this constant and we also haven't written W as a function of T just right yet it's kind of a they're their input implicitly expressed this is more of a relationship than a function right now so the first thing we can do is try to solve for C we're told in the problem we're told in the problem that W of 0 is equal to W of 0 is equal to 1,400 tons they told us that that's one of our initial conditions so we know that the natural log that when T is equal to 0 W is equal to 1,400 so 1400 W is equal to 1,400 when t is equal to 0 is equal to 0 plus C or another way to think about it is the natural log of the absolute value of 1100 is equal to C and obviously this is a positive value so we can just say that C is equal to the natural log of 1,100 we can drop the absolute value signs because the absolute value of 11 or it is 1,100 so now we can rewrite this part right over here let me do it in yellow let me start that's not yellow let me start right over here so we could write the natural log of the absolute value of W minus 300 and actually we know that W is an increasing function W starts at 1,400 and only goes up from there as T gets larger and larger so if we're starting at 1400 and only getting larger and larger values for W this expression right here is never going to get is never going to get negative 4 positive T values and that's all we care about and so we can actually drop the absolute value sign this will always be a positive value so we could say the absolute value of W minus 300 is equal to T over 25 and we've just solved for our constant plus the natural log of 1,100 and now we can just solve this explicitly for W we can raise we can raise e to both sides to this value is equal to this value the left-hand side is equal to the right hand side so earase to this power is the same thing as e raised to that power so we have e to the natural log of W minus 300 is going to be equal to e to the t over 25 plus the natural log of 1,100 and let me let me take this let me actually just to get the real estate let me delete that alright now I can keep working right below that and so e to the natural log of something is just going to be equal to that something the natural log is what power do I have to raise e to to get to get this thing well I'm raising you to that power so I'm going to get this thing so this the left hand side just becomes W minus 300 and the right hand side we could rewrite this as e to the t over 25 times e to the natural log of 1,100 all right we have the same base different exponents you would if you wanted to simplify this you would just add the exponents which is exactly what we did up here so these two things are equivalent e to the natural log of 1,100 that is 1,100 the natural log of 11 our is a power that you raise e to to get to 1,100 so if you raise e to that power you get 1,100 so that is 1,100 so we get W minus 300 is equal to 1,100 1,100 e to the t over 25 and then you just add 300 to both sides w is equal to 1,100 e to the t over 25 plus 300 and we're done we have solved the differential equations we found the particular solution we had to solve for C to find that particular solution but we're done and we did it using just basic integration we didn't have to use any of the fancy yeah differential equation solving tools that you might learn in a more advanced class

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